Is Every Element of n-Torus DeRham Cohomology Independent?

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Discussion Overview

The discussion centers on the independence of elements in the deRham cohomology of the n-torus, specifically regarding the computation of the m-th cohomology group. Participants explore theoretical aspects, including dimension arguments and relationships between homology and cohomology groups.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant claims to have shown that each form [dθi_1 ∧ dθi_2 ... ∧ dθi_m] is independent in the deRham cohomology, suggesting that these forms are closed but not exact.
  • The same participant argues that the dimension of the m-th cohomology group is at least C(n,m) based on their computations and reasoning involving CW-complexes.
  • There is a claim that the dimension of the m-th cohomology group is also C(n,m), linking it to the dimensions of homology groups without needing to compute them directly.
  • Another participant questions the initial post by asking why the Kunneth formula with the reals as the coefficient group is not being utilized in the discussion.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as one participant's reasoning is challenged by another, indicating that there are competing views on the approach to the problem.

Contextual Notes

The discussion includes assumptions about the relationships between cellular homology, simplicial homology, and deRham cohomology, as well as the application of the Kunneth formula, which remain unresolved.

Sina
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Hello,

I want to ask for a verification of something I did. Let's say I want to compute the m-th cohomology group of the n-torus with coordinates θi. Suppose that I have shown each [dθi_1 \wedgedθi_2 ... \wedge dθi_m] is an independent element of this deRham cohomology (it can be shown by some integration over m-toruses that reside in the n-torus that these are closed but not exact differential forms whose differences are not exact). In particular its dimension is then atleast C(n,m)

Now to finish I have to show that this is all. I want to a dimension argument here. In the lattice quotient model of the Torus it is easy to show that each n-torus is a CW-complex with 1 0-cell, n 1-cells, C(n,m) m-cells. Then one can compute the homology groups and dimension of its mth homology group is C(n,m) then dimension of the mth cohomology group is also m (I know the computation of homology groups for T2 is easy I assume Tn is the similar). In fact you don't need to compute the homology groups. Since each chain Cm has dimension C(n,m) then the homology group can have at most that dimension which is more than enough for me.

Now we now that for simplical complices Hm(X) = Hm(Xk) for k>m. In particular here we have that Xn is the Torus itself so that cellular homology and normal homology groups can be identifies (this is the first part I am not sure). Then normal homology group is isomorphic to simplical homology groups and there fore each have the same dimension. Now simplical cohomology groups of simplical complexes are isomorphic to deRham cohomology. Thus basically m-th deRham cohomology group has the same dimension as the mth cohomology of the n-torus which is C(n,m) therefore completing the proof.

Right? :p
 
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edit: finally manged to tex
 
Last edited:
I meant " then dimension of the mth cohomology group is also C(n,m)" not m
 
i have read this a couple of times am am not sure what you are asking.

For instance why are you not using the Kunneth formula with the reals as the coefficient group?
 

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